# Thread: Correct me if im wrong.. but

1. ## Correct me if im wrong.. but

The function of $cos \pi/6 =$ the cofunction of $sin \pi/6$ or something similar... (forgot how it was worded)

and would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x)

2. Originally Posted by 3deltat
The function of $cos \pi/6 =$ the cofunction of $sin \pi/6$ or something similar... (forgot how it was worded)

and would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x)
Sine and Cosine are Co-functions.
Meaning,
$\sin x=\cos \left( \frac{\pi}{2} -x \right)$
In the same way,
Tangent and Cotangent are cofunctions.
And,
Secant and Cosecant are cofunctions.

You can use this mnemonic to remember the properties of trigonometric cofunctions.

3. Sine and Cosine are Co-functions.
Meaning,

In the same way,
Tangent and Cotangent are cofunctions.
And,
Secant and Cosecant are cofunctions.
Ah. So the statement "The function of $cos \pi/6$ Would = the cofunction of $sin \pi/6$ ? =)

4. Originally Posted by 3deltat
Ah. So the statement "The function of $cos \pi/6$ Would = the cofunction of $sin \pi/6$ ? =)
Thus, the by the cofunction identities,
$\cos \frac{\pi}{6}=\sin \left( \frac{\pi}{2} - \frac{\pi}{6} \right)=\sin \frac{\pi}{3}$

5. Originally Posted by 3deltat
would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x)
As sine is an odd function $sin(-x) = -sin(x)$. Thus
$-2sin(4x)sin(-2x) = -2sin(4x) \cdot -sin(2x) = 2sin(4x)sin(2x)$.